f) \( (3 x+5)(3 x-5)= \) g) \( (2 x-4)(2 x+4)= \) h) \( (x+2)^{3}= \) i) \( (x-4)^{3}= \)
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Bonus Knowledge
To solve these expressions, we can use the identities for the difference of squares and the binomial expansion. For the first expression, \( (3 x+5)(3 x-5) \) is in the form \( (a+b)(a-b) \), which simplifies to \( a^2 - b^2 \). So, it becomes \( (3x)^2 - (5)^2 = 9x^2 - 25 \). The second expression, \( (2 x-4)(2 x+4) \), follows the same pattern: \( (a-b)(a+b) \). Here, \( a = 2x \) and \( b = 4 \), so it simplifies to \( (2x)^2 - (4)^2 = 4x^2 - 16 \). For \( (x+2)^3 \), we can use the binomial theorem or the formula for cubes: \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). Thus, it becomes \( x^3 + 3(2)(x^2) + 3(2^2)(x) + 2^3 = x^3 + 6x^2 + 12x + 8 \). Finally, for \( (x-4)^3 \), it follows the same principle: \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \). So, it simplifies to \( x^3 - 3(4)(x^2) + 3(4^2)(x) - 4^3 = x^3 - 12x^2 + 48x - 64 \). Now you have the simplified forms: f) \( 9x^2 - 25 \) g) \( 4x^2 - 16 \) h) \( x^3 + 6x^2 + 12x + 8 \) i) \( x^3 - 12x^2 + 48x - 64 \)